--%>

Schrodinger equation with particle in a box problem.

Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.

So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functions for a particle that is free to move in three dimensions are deduced. A molecule of a gas enclosed in a cubic container provides a specific example that is dealt with in the section after the general procedure has been developed.

For any three-dimensional problem, the potential energy is, general, a function of three coordinates. For a cubic potential box, the Cartesian coordinates are convenient. The differential equation that must be solved is now the Schrodinger equation in three dimensions.

1310_Particle in a box.png 

For a "cubic box," the potential function can be expressed in terms of separate x, y, and z components,
98_Particle in a box1.png 

Each of the potential function components for a "particle-in-a-box" is like the one-dimensional potential for a "particle-on-a-line".

For three-dimensional systems, the solution function ψ depends on the three coordinates necessary to locate a point in space. It is often profitable to try to separate such systems into parts, with each part involving only one coordinate. On the basis we try the substitution

ψ (x, y, z) = Ø(x)Ø(y)Ø(z)

Substitution of (2) from (1) gives

1350_Particle in a box2.png 

Division by Ø(x)Ø(y)Ø(z) gives

1180_Particle in a box3.png 

For the equation to be satisfied for all values of x, y and z, each term on the left must equal a component of ε, and we can write

ε = εx + εy + εz

The Schrodinger equation can then be broken down into three identical equations of the type

1394_Particle in a box4.png 

Or

578_Particle in a box6.png 

These equations are identical to that written for one-dimensional problem. The solution to the three-dimensional cubic-box problem is therefore

ψ =  Ø(x)Ø(y)Ø(z)

With

1809_Particle in a box7.png

   Related Questions in Chemistry

  • Q : Molarity based question Help me to

    Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M

  • Q : What are homogenous catalyst? Give few

    When a catalyst mixes homogeneously with the reactants and forms a single phase, the catalyst is said to be homogeneous and this type of catalysis is called homogeneous catalysis. Some more examples of homogeneous catalysis are:    SO2

  • Q : Problem on Neutralization What weight

    What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution? (a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g      Answer

  • Q : Cations Explain how dissolving the

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Problem on equilibrium constant Ethanol

    Ethanol is manufactured from carbon monoxide and hydrogen at 600 K and 20 bars according to the reaction2 C0(g) + 4 H2(g) ↔ C2H5OH(g) + H2O (g)The feed stream contains 60 mol% H2, 20 m

  • Q : Questuion associated with colligative

    Provide solution of this question. Which of the following is a colligative property: (a) Surface tension (b) Viscosity (c) Osmotic pressure (d) Optical rotation

  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

  • Q : Describe Thermodynamics Properties The

    The free energy property leads to convenient expressions for the volume and pressure dependence of internal energy, enthalpy and the heat capacities.All the properties of a chemical system, a sample of a substance, or a mixture of substances have some fixe

  • Q : Benzoic acid is weaker than paranitro

    Briefly state that Benzoic acid is weaker than paranitro benzoic acid?

  • Q : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (